A) 5
B) 2
C) 1
D) 0
Correct Answer: D
Solution :
First time \[=\frac{2}{\sqrt{7}+\sqrt{5}}\] \[=\frac{2\times (\sqrt{7}-\sqrt{5})}{(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})}\] \[=\frac{2\times (\sqrt{7}-\sqrt{5})}{7-5}=\sqrt{7}-\sqrt{5}\] Second time \[=\frac{7}{\sqrt{12}-\sqrt{5}}\] \[=\frac{7(\sqrt{12}+\sqrt{5})}{(\sqrt{12}-\sqrt{5})(\sqrt{12}+\sqrt{5})}\] \[=\frac{7(\sqrt{12}+\sqrt{5})}{12-5}\] \[=\frac{7(\sqrt{12}+\sqrt{5})}{7}=\sqrt{12}+\sqrt{15}\] Third time \[=\frac{5}{\sqrt{12}-\sqrt{7}}\] \[=\frac{5(\sqrt{12}+\sqrt{7})}{(\sqrt{12}-\sqrt{7})(\sqrt{12}+\sqrt{7})}\] \[=\frac{5(\sqrt{12}+\sqrt{7})}{12-7}=(\sqrt{12}+\sqrt{7})\] \[\therefore \] Expression \[=(\sqrt{7}-\sqrt{5})+(\sqrt{12}+\sqrt{5})-(\sqrt{12}+\sqrt{7})\] \[=\sqrt{7}-\sqrt{5}+(\sqrt{12}+\sqrt{5})-(\sqrt{12}-\sqrt{7})=0\]You need to login to perform this action.
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